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Of course, it does not oscillate. Here, the voltage of capacitor is fixed to the supply voltage and the current of inductor rises linearly with time: Ldi/dt=Vdd ==> i=(Vdd/L)t. therefore, it produce magnetic field and even even electric field (E prop. dB/dt). Because the capacitor is in parallel...
In the way that you have connected the input, there is no feedback: in the small signal analysis, all the independent voltage source will be grounded as a result the base node of Q1 will be shorted to ground and therefore, there is no feedback :).
The output impedance in this case will be...
It is not about RC. Every system that has a bandwidth limitation distorts the pulse shape: It can be RC, fiber optics or anything-else. The bandwidth limitation can be due to multi-path fading. See the attachments.
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This means that the order of the channel is higher than...
1) to make the DC voltage at node 3 equal to R2/(R2+R1)*12~0.5 V
2) The capacitor C4 is used to make this node ground at high frequency.
"Did you mean that C4 charging is done through the parallel combination of R1 and R2, and discharging through R2? "
No, it will be charged and discharged by...
Increase the size of transistors till you get osicllation. ( i assume that you already know how to simulate an oscillator.) you may wonder why you need to increase the size of transistors to achieve an oscillation. It is needed to increase the oscillation period to be much larger than the rise...
Of course Rd should add noise to the output even when the input is open. However, the noise contribution of M1 is zero, because the current noise of M1, due to open input port, can not flow into the output.
Just count the number of reactive components. Let us call it A. Then count the number of capacitive loops and inductive cut sets in the circuit B and C respectively. Then the order of circuit is A-B-C. Very easy. For convenience, i have attached an example.
The time constant at node 3 is determined by the parallel combination of R1 and R2 and C4. So its time constant is much smaller than what you have stated.
The book does not say that. Just wants to emphasize on the fact that the bias current adds to the overall noise of the circuit, which is very important in low voltage design( why?).
First of all in differential circuits, the differential output voltage gain is 6dB larger than each single ended output.
38 --> 44
If your circuit is fully differential, the phase-margin can be found from either by differential output or single ended ones. Both shoul give you the same phase...
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